Squares and cubes problem

By Maria Jose Leon-Sotelo Esteban (T3819) on Friday, May 11, 2001 - 07:18 am :

Find two numbers,such that the sum of their cubes is a square and the sum of their squares a cube.
Thank you
Maria Jose.

By David Loeffler (P865) on Friday, May 11, 2001 - 11:08 am :

2 and 2 will do. Did you want distinct numbers?

By David Loeffler (P865) on Friday, May 11, 2001 - 11:15 am :

(In fact, if both are 2^6k+1 for any k, we get a solution.)

By David Loeffler (P865) on Friday, May 11, 2001 - 11:19 am :

However, I have found no other solutions so far, and I have done a quick computer check on all pairs of integers both less than 150.

I'll get back to you if I find out anything more.

David

By David Loeffler (P865) on Friday, May 11, 2001 - 01:40 pm :

There are, in fact, infinitely many solutions for every value of the ratio of the two.

Here is a construction for solutions:

Suppose that x, y are the two integers, and x/y = p/q is known. Then we have x = pw, y = qw, and we must solve

(p² + q² )w² = a³

(p³ + q³ )w³ = b²

Let w = (p² + q² )^6j+4 (p³ + q³ )^6k+3 .

Try it; it works, for any non-negative integers j and k. We obtain
a³ = (p² + q² )^12j+9 (p³ + q³ )^12k+6 , so a = (p² + q² )^4j+3 (p³ + q³ )^4k+2 ,
and b² = (p² + q² )^18j+12 (p³ + q³ )^18k+10 , so b = (p² + q² )^9j+6 (p³ + q³ )^9j+5 .

For example, in the case p = 2, q = 1 we get w = 5^6k+4 9^6k+3 , so taking j = k = 0 we have w = 455625. So x = 911250, y = 455625 works.

(There are other solutions. For example, 625 and 1250 work. Can anyone think of a general form for all solutions?)

By Maria Jose Leon-Sotelo Esteban (T3819) on Friday, May 11, 2001 - 02:50 pm :

Thank you.
Inpressive and incredible answer!!!
Maria Jose.

By Michael Doré (Michael) on Friday, May 11, 2001 - 06:52 pm:

Nice construction, David!

By Brad Rodgers (P1930) on Saturday, May 12, 2001 - 04:17 am :

I have an idea that seems to show some promise for finding all pairs, but I can't seem to finish it.

First, note that

n-1
å
j=0

3j²+3j+1=n³

So for a² +b² =n³

We may just add 3j² +3j+1 from 0 until we come up with a perfect square, calling this perfect square a² . Note that j was i-1 at this point, and that i is necessarily a perfect square. Now sum from i to some number, calling this number n-1, until we get another perfect square, calling this perfect square b² . We now know that a² +b² =n³

A similar method would work using odd numbers to get the equation

a³ +b³ =m²

Perhaps someone else can do something useful with this (like finding a pythagorean-triple-like algorithm...)

Brad

By David Loeffler (P865) on Monday, May 14, 2001 - 01:23 pm :

Brad,

How will this generate all pairs? It looks to me like we would have to have a being a sixth power for this to work.
(Sorry if I am misinterpreting your post.)

David

By Brad Rodgers (P1930) on Monday, May 14, 2001 - 05:56 pm :

Actually, I don't think it will generate all pairs. It will generate all possible a's though. I'm not sure if I understand though what you are saying "we would have to have a being a sixth power for this to work."

Brad

By Brad Rodgers (P1930) on Monday, May 14, 2001 - 07:43 pm :

Ok, I see what you are saying now. You're right, using my method, the number a would have to be a sixth power.

By Brad Rodgers (P1930) on Monday, May 14, 2001 - 11:09 pm :

Or, rather, a² would need to be a sixth power.

Brad

By David Loeffler (P865) on Tuesday, May 15, 2001 - 01:43 pm :

Yeah, sorry, that is what I meant.